What is the current in a wire of radius R = 2.67 mm if the magnitude of the current density is given by (a) `J_n= J_0 r//R and (b) J_b=J_0(1-r//R),` in which r is the radial distance and `J_0= 5.50 xx 10^(4) A//m^(2)?` (c) Which function maximizes the current density near the wire's surface?

Answer

To solve the problem, we will break it down into three parts as described in the question.

### Given Data:
- Radius of the wire, \( R = 2.67 \, \text{mm} = 2.67 \times 10^{-3} \, \text{m} \)
- Current density for part (a): \( J_n = J_0 \frac{r}{R} \)
- Current density for part (b): \( J_b = J_0 \left(1 - \frac{r}{R}\right) \)
- \( J_0 = 5.50 \times 10^{4} \, \text{A/m}^2 \)

### Part (a): Calculate the Current \( I_n \)

1. **Expression for Current Density**:
\[ J_n = J_0 \frac{r}{R} \]

2. **Differential Area Element**:
The differential area element \( dA \) in cylindrical coordinates is given by:
\[ dA = 2\pi r \, dr \]

3. **Total Current**:
The total current \( I_n \) can be calculated by integrating the current density over the cross-sectional area of the wire:
\[ I_n = \int_0^R J_n \, dA = \int_0^R J_0 \frac{r}{R} (2\pi r \, dr) \]

4. **Substituting and Simplifying**:
\[ I_n = \int_0^R J_0 \frac{r}{R} (2\pi r \, dr) = \frac{2\pi J_0}{R} \int_0^R r^2 \, dr \]
The integral \( \int_0^R r^2 \, dr = \frac{R^3}{3} \).

5. **Final Expression for Current**:
\[ I_n = \frac{2\pi J_0}{R} \cdot \frac{R^3}{3} = \frac{2\pi J_0 R^2}{3} \]

6. **Substituting Values**:
\[ I_n = \frac{2\pi (5.50 \times 10^4) (2.67 \times 10^{-3})^2}{3} \]
\[ I_n = \frac{2\pi (5.50 \times 10^4) (7.1289 \times 10^{-6})}{3} \]
\[ I_n \approx 0.821 \, \text{A} \]

### Part (b): Calculate the Current \( I_b \)

1. **Expression for Current Density**:
\[ J_b = J_0 \left(1 - \frac{r}{R}\right) \]

2. **Total Current**:
The total current \( I_b \) can be calculated similarly:
\[ I_b = \int_0^R J_b \, dA = \int_0^R J_0 \left(1 - \frac{r}{R}\right) (2\pi r \, dr) \]

3. **Substituting and Simplifying**:
\[ I_b = \int_0^R J_0 \left(1 - \frac{r}{R}\right) (2\pi r \, dr) = 2\pi J_0 \left( \int_0^R r \, dr - \frac{1}{R} \int_0^R r^2 \, dr \right) \]
The integrals yield:
\[ \int_0^R r \, dr = \frac{R^2}{2}, \quad \int_0^R r^2 \, dr = \frac{R^3}{3} \]

4. **Final Expression for Current**:
\[ I_b = 2\pi J_0 \left( \frac{R^2}{2} - \frac{1}{R} \cdot \frac{R^3}{3} \right) = 2\pi J_0 \left( \frac{R^2}{2} - \frac{R^2}{3} \right) \]
\[ I_b = 2\pi J_0 \cdot \frac{3R^2 - 2R^2}{6} = \frac{2\pi J_0 R^2}{6} = \frac{\pi J_0 R^2}{3} \]

5. **Substituting Values**:
\[ I_b = \frac{\pi (5.50 \times 10^4) (2.67 \times 10^{-3})^2}{3} \]
\[ I_b \approx 0.411 \, \text{A} \]

### Part (c): Determine Which Function Maximizes Current Density Near the Surface

- From the calculations:
- \( I_n \approx 0.821 \, \text{A} \)
- \( I_b \approx 0.411 \, \text{A} \)

Since \( I_n > I_b \), the function \( J_n = J_0 \frac{r}{R} \) maximizes the current density near the wire's surface.

### Summary of Results:
- (a) Current \( I_n \approx 0.821 \, \text{A} \)
- (b) Current \( I_b \approx 0.411 \, \text{A} \)
- (c) The function that maximizes the current density near the wire's surface is \( J_n = J_0 \frac{r}{R} \).